## anonymous 5 years ago Gasoline is pouring into a cylindrical tank of radius 3 feet. When the depth of the gasoline is 6 feet, the depth is increasing at 0.2 ft/sec. How fast is the volume of gasoline changing at that instant? Round your answer to three decimal places.

1. anonymous

Okay, you can find the volume first since you have h and r , so : $V = \pi r^2 h$ $= ( 3)^2 (6) \pi$ = 169.65 ft^3

2. anonymous

sstarica, isn't this the same thing we were tackling

3. anonymous

now find V' which is :$V' = 2\pi rr'h'$

4. anonymous

not really @iam , somehow similar :)

5. anonymous

now we have: h' = 0.2 ft/sec we need to find r'

6. anonymous

do i just plug in

7. anonymous

yes :) but we need to find r' now

8. anonymous

$V' = 2 \pi rr' h'$ we have h' , but we need to find r' to calculate V' = rate of volume increase

9. anonymous

how do i do that? :(

10. anonymous

hold on :)

11. anonymous

thx

12. anonymous

you need to find a relationship between r and h so r/h = r'/h'

13. anonymous

thats where im having problem for all the problems

14. anonymous

cross mulitply and you'll get: r' = rh'/r

15. anonymous

sorry! wait you'll get: r' = rh'/h :)

16. anonymous

are you with me soo far?

17. anonymous

yes :)

18. anonymous

now plug : r =3 ft h = 6ft h' = 0.2ft/sec r' = (3)(0.2)/6 = 0.1 ft/sec

19. anonymous

we have found r' now, let's go back to V'$V' = 2 \pi rr'h'$ so, V' = 2(3)(0.1)(0.2) pi = 0.377 ft^3/sec

20. anonymous

Correct me if I'm wrong please

21. anonymous

OMG this makes so much sense now

22. anonymous

but the website marked it wrong :(

23. anonymous

why?

24. anonymous

I guess there's a mistake, but that's the idea, maybe in calculation?

25. anonymous

ok

26. anonymous

did you get the idea?

27. anonymous

one qquestion

28. anonymous

sure :)

29. anonymous

where did u get $V \prime = 2\pi r r \prime h \prime$

30. anonymous

you derive with respect to r and h :)

31. anonymous

you want to know how is the volume increasing when both the radius and height is changing, that's the story of the related rated problem

32. anonymous

oooooh...im sorry can u show me one more problem

33. anonymous

hmmmm, let's say we have a circle of radius = 2 m, and the radius is decreasing at 0.5 m/sec. How much will the circle's area increase ?

34. anonymous

I'll give you advise, whenever you see ( m/sec) or anything with respect to time it's something' which in this case is the r' = 0.5m/sec

35. anonymous

so what is the given?

36. anonymous

we have : - R = 2 m - R' = 0.5 m/sec and we know the formula of the Area of the circle which is: A = pi r^2, right?

37. anonymous

uhhuh

38. anonymous

Now , you want to know how much is the area increasing with respect to R, since R is increasing :) so you find the derivative of it which will be : A' = 2 pi rr'

39. anonymous

so when exactly do i start pluging in numbers?

40. anonymous

right now, after you find the derived formula, start pluggingthe numbers :)

41. anonymous

okay ...these are pretty simple compared to the ones i have on hw :'(

42. anonymous

Like this problem, they want to know how much is the area increasing , we got the formula so the answer is: A' = 2(2)(0.5) pi = 2 pi = 2.68 m^2/sec :) So the area is increasing at 2.68 m^2/sec

43. anonymous

Lol, when you understand the small idea, you'll be able to figure out the bigger idea :)

44. anonymous

did you understand my problem?

45. anonymous

yes but how would i approach towards a hemisphere problem

46. anonymous

think about it, write down the given on one side, and the RTF(required to find) on the other side, then SKETCH! To test your understanding , you've got to sketch what they give you to be fully understand what they want :)

47. anonymous

to fully*

48. anonymous

1) write the given and what you need to find 2) sketch 3) write down the general formula that you're going to use :)

49. anonymous

^_^

50. anonymous

ok :)

51. anonymous

good luck ! You're a smart student, you can figure it out :)

52. anonymous

you got the small idea, you'll be able to solve any bigger ideas now . They are not even big, they are small ideas that got stretched ^_^, don't let the problem boss u, YOU ARE THE BOSS!

53. anonymous

lol..thx

54. anonymous

np :) Give it a try once more. ^_^